(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A$ is the area of the plates and $d$ is the dis

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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.When the distance $d$ is increased,the capacitance $C$ decreases.Since the battery is disconnected,the charge $Q$ on the plates remains constant.The relationship between charge,capacitance,and potential difference is $V = \frac{Q}{C}$.Since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.

Class 12 Physics Electric Potential and Capacitance Parallel Plate Capacitor

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$A$ parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates:

AIPMT 2006 All AIPMT

A
decreases
B
increases
C
does not change
D
becomes zero

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Class 12 Questions

Class 12

Physics Questions

Chapter

Electric Potential and Capacitance

Subtopic

Parallel Plate Capacitor

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Two parallel air capacitors have their plate areas $100 \,cm^2$ and $500 \,cm^2$, respectively. They have the same charge and potential. If the distance between the plates of the first capacitor is $0.5 \,mm$, the distance between the plates of the second capacitor is (in $\,cm$)

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Six metallic plates,each with a surface area $A$ on one side,are placed at a distance $d$ from each other. The alternate plates are connected to points $P$ and $Q$ as shown in the figure. The capacitance of the system is:

Medium

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The following arrangement consists of five identical metal plates parallel to each other. The area of each plate is $A$ and the separation between successive plates is $d$. The capacitance between $P$ and $Q$ is

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$A$ metal slab of thickness $b = d/2$ is inserted between the plates of a parallel plate capacitor of plate separation $d$. What is the ratio of the new capacitance to the original capacitance?

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The plates of a parallel plate capacitor of capacity $50\,\mu F$ are charged to a potential of $100\;V$ and then disconnected from the battery. The distance between the plates is then doubled. How much work is done in doing so?

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$A$ parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates:

Similar Questions

Two parallel air capacitors have their plate areas $100 \,cm^2$ and $500 \,cm^2$, respectively. They have the same charge and potential. If the distance between the plates of the first capacitor is $0.5 \,mm$, the distance between the plates of the second capacitor is (in $\,cm$)

Six metallic plates,each with a surface area $A$ on one side,are placed at a distance $d$ from each other. The alternate plates are connected to points $P$ and $Q$ as shown in the figure. The capacitance of the system is:

The following arrangement consists of five identical metal plates parallel to each other. The area of each plate is $A$ and the separation between successive plates is $d$. The capacitance between $P$ and $Q$ is

$A$ metal slab of thickness $b = d/2$ is inserted between the plates of a parallel plate capacitor of plate separation $d$. What is the ratio of the new capacitance to the original capacitance?

The plates of a parallel plate capacitor of capacity $50\,\mu F$ are charged to a potential of $100\;V$ and then disconnected from the battery. The distance between the plates is then doubled. How much work is done in doing so?

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